cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A003320 a(n) = max_{k=0..n} k^(n-k).

Original entry on oeis.org

1, 1, 1, 2, 4, 9, 27, 81, 256, 1024, 4096, 16384, 78125, 390625, 1953125, 10077696, 60466176, 362797056, 2176782336, 13841287201, 96889010407, 678223072849, 4747561509943, 35184372088832, 281474976710656, 2251799813685248
Offset: 0

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Author

Keywords

Comments

For n > 0, a(n+1) = largest term of row n in triangles A051129 and A247358. - Reinhard Zumkeller, Sep 14 2014

Examples

			a(5) = max(5^0, 4^1, 3^2, 2^3, 1^4, 0^5) = max(1,4,9,8,1,0) = 9.
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • I. Tomescu, Introducere in Combinatorica. Editura Tehnica, Bucharest, 1972, p. 231.

Crossrefs

Programs

  • Haskell
    a003320 n = maximum $ zipWith (^) [0 .. n] [n, n-1 ..]
    -- Reinhard Zumkeller, Jun 24 2013
    
  • Mathematica
    Join[{1},Max[#]&/@Table[k^(n-k),{n,25},{k,n}]] (* Harvey P. Dale, Jun 20 2011 *)
  • PARI
    a(n) = vecmax(vector(n+1, k, (k-1)^(n-k+1))); \\ Michel Marcus, Jun 13 2017

Formula

a(n) = A056155(n-1)^(n - A056155(n-1)), for n >= 2. - Ridouane Oudra, Dec 09 2020

Extensions

Easdown reference from Michail Kats (KatsMM(AT)info.sgu.ru)
More terms from James Sellers, Aug 21 2000

A265899 After a(1) = 1, positions of descents in A265894.

Original entry on oeis.org

1, 3, 6, 8, 11, 14, 17, 20, 24, 27, 31, 34, 38, 41, 45, 49, 53, 57, 61, 65, 69, 73, 77, 81, 85, 89, 94, 98, 102, 106, 111, 115, 120, 124, 128, 133, 137, 142, 146, 151, 156, 160, 165, 169, 174, 179, 184, 188, 193, 198, 202, 207, 212, 217, 222, 227, 231, 236, 241, 246, 251, 256, 261, 266, 271, 276, 281, 286, 291, 296, 301
Offset: 1

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Author

Antti Karttunen, Dec 24 2015

Keywords

Comments

Numbers n for which A099563(A001813(n)) <= A099563(A001813(n-1)), where A001813(n) = (2n)! / n!, and A099563 gives the most significant digit in the factorial base representation (A007623) of n.

Crossrefs

Cf. A265898 (a subsequence), A266119 (first differences), A266120 (terms immediately before descents).
Cf. also A031435.

Programs

  • PARI
    A099563(n) = { my(i=2,dig=0); until(0==n, dig = n % i; n = (n - dig)/i; i++); return(dig); };
    A265894 = n -> A099563((2*n)! / n!);
    my(i=0, p=1, n=0); while(i < 60, n++; my(k = A265894(n)); if(k <= p, i++; print1(n, ", ")); p = k; );
    
  • Scheme
    ;; With Antti Karttunen's IntSeq-library.
    (define A265899 (MATCHING-POS 1 1 (lambda (n) (<= (A265894 n) (A265894 (- n 1))))))

A307175 Smallest power to which 1+1/n must be raised in order for an interval [k,k+1], with k an integer, to be skipped.

Original entry on oeis.org

4, 7, 9, 13, 15, 18, 22, 26, 27, 32, 33, 40, 42, 48, 51, 55, 58, 62, 66, 71, 75, 80, 85, 85, 91, 97, 103, 105, 111, 112, 120, 121, 129, 131, 139, 142, 143, 153, 156, 158, 168, 172, 175, 178, 181, 193, 197, 201, 206, 210, 215, 220, 225, 230, 235, 241, 246, 252
Offset: 2

Views

Author

Alex Costea, Mar 27 2019

Keywords

Comments

Here the skipping of an interval means that the interval falls strictly between (1+1/n)^(a(n)-1) and (1+1/n)^a(n).
The sequence is not monotonically increasing; a(24) = a(25) and a(62) > a(63) are the first counterexamples.
Asymptotic to n * log(n), and as such also to the prime numbers (A000040).

Examples

			1.1^26 = 11.918... and 1.1^27 = 13.109...; [12,13] is skipped, and this is the first time this happens, thus a(10)=27.
		

Crossrefs

Cf. A031435.

Programs

  • Mathematica
    a[n_, m_] := Reduce[(1+1/n)^(m-1) < k < k+1 < (1+1/n)^m, k, Integers];
    a[n_] := For[m = 1, True, m++, If[a[n, m] =!= False, Return[m]]];
    Table[a[n], {n, 2, 100}] (* Jean-François Alcover, Jul 07 2019 *)
  • PARI
    a(n) = my(k=2, last=1+1/n); while(floor(new = (1+1/n)^k) - ceil(last) != 1, k++; last = new); k; \\ Michel Marcus, Mar 30 2019
    
  • Python
    from math import floor, log
    def get_a_of_n(i):
         x=1+1/i
         j=i
         while floor(log(j, x))!=floor(log(j+1, x)):
             j+=1
         return floor(log(j, x))+1
    def main():
         step=1
         i=2
         while True:
             y=get_a_of_n(i)
             print(y, end=", ")
             i+=step

A377206 a(n) = ceiling(log(1/n)/log(1 - 1/n)).

Original entry on oeis.org

1, 3, 5, 8, 10, 13, 16, 19, 22, 26, 29, 33, 36, 40, 43, 47, 51, 55, 59, 63, 67, 71, 75, 79, 84, 88, 92, 96, 101, 105, 110, 114, 119, 123, 128, 132, 137, 142, 146, 151, 156, 160, 165, 170, 175, 180, 184, 189, 194, 199, 204, 209, 214, 219, 224, 229, 234
Offset: 2

Views

Author

Walter Robinson, Oct 19 2024

Keywords

Comments

This sequence also describes the minimum number of n-player games, where each player has an equal chance of winning, that must be played for a given player to have an equal or greater chance of winning at least once than they have of losing a single game.

Crossrefs

Programs

  • Mathematica
    Table[Ceiling[Log[1/n]/Log[1-1/n]], {n,2,58}] (* James C. McMahon, Nov 04 2024 *)

Formula

a(n) = A031435(n-1) + 1 for n >= 3.
Showing 1-4 of 4 results.